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Question 1
ప్రశ్న 1
Which statement best describes an ideal perfectly flexible cable used in elementary structural analysis?
Explanation:
An ideal flexible cable is assumed to possess negligible flexural stiffness. It therefore cannot sustain bending moment in the elementary cable model. It also cannot carry axial compression because a flexible cable would simply lose tension or change configuration. Its internal force is axial tension directed along the local tangent to the cable. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which comparison among an ideal cable, a beam and an arch is correct?
Explanation:
A flexible cable adapts its geometry so that its principal internal action is axial tension. An arch shaped funicularly for its loading carries the corresponding load primarily through axial compression. A conventional beam subjected to transverse loading develops bending moments and shear forces. These different load-transfer mechanisms are fundamental distinctions among the three structural forms. Therefore Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which statement correctly identifies the functions of major components of a conventional suspension bridge?
Explanation:
The suspended deck transfers vertical load to the main cables through hangers or suspenders. The main cables pass over or are supported by the towers. Large cable tensile forces must ultimately be resisted at the ends, commonly by substantial anchorages in an externally anchored suspension bridge. The towers primarily transmit large vertical components of cable force toward the foundations. Therefore Option D correctly describes the principal load path.
వివరణ:
Question 4
ప్రశ్న 4
Which statement correctly distinguishes two standard ideal cable shapes?
Explanation:
A constant vertical load intensity measured per unit horizontal projection leads to a parabolic cable equation. This loading idealization is widely used for suspension-bridge dead load transferred through closely spaced hangers. Cable self-weight, however, is naturally distributed per unit length measured along the curved cable. The exact equilibrium shape for that self-weight condition is a catenary rather than a parabola. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
For a perfectly flexible cable supported at the same level and subjected only to vertical loads, let H be the constant horizontal component of cable tension and y be the vertical dip below the support chord at a section. If M0 is the bending moment at the corresponding section of a simply supported beam carrying the same vertical loads, which relation expresses the general cable theorem?
Explanation:
The corresponding simply supported beam develops a bending moment M0 under the specified vertical loads. The flexible cable itself cannot sustain bending moment. The horizontal component H acting through the cable dip y provides the balancing moment Hy. Equilibrium therefore gives Hy = M0 at the corresponding section. Hence Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
A perfectly flexible cable has supports at the same level, span L and midspan sag f. It carries a uniformly distributed vertical load w per unit horizontal span. Which pair correctly gives its parabolic profile and constant horizontal component of tension?
Explanation:
The cable passes through the two supports, so its dip must be zero at x = 0 and x = L. Its maximum dip f occurs at midspan. The parabolic equation satisfying these conditions is y = 4fx(L-x)/L². Applying the general cable theorem at midspan gives Hf = wL²/8, so H = wL²/(8f). Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
A flexible cable has supports at the same level, a span of 24 m and a midspan sag of 3 m. It carries a UDL of 8 kN/m over the horizontal span. What are the vertical reaction at each support and the horizontal component H of cable tension?
Explanation:
The total vertical load is 8 × 24 = 192 kN. Symmetry therefore gives a vertical reaction of 96 kN at each support. For a parabolic cable, H = wL²/(8f). Substitution gives H = 8 × 24²/(8 × 3) = 192 kN. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
For the cable in the previous question, what are the minimum cable tension and the maximum cable tension approximately?
Explanation:
At midspan the cable tangent is horizontal, so the cable tension equals its horizontal component H. Thus the minimum tension is 192 kN. At either support, the tension has horizontal component 192 kN and vertical component 96 kN. Its magnitude is sqrt(192² + 96²), which is approximately 214.7 kN. Therefore Option D is correct.
వివరణ:
Question 9
ప్రశ్న 9
For the 24 m span, 3 m sag cable carrying 8 kN/m, what are the cable dip and tension magnitude at x = 6 m from the left support?
Explanation:
The parabolic ordinate is y = 4(3)(6)(18)/24² = 2.25 m. The vertical-force component at that section equals the corresponding beam shear, 96 - 8(6) = 48 kN. The horizontal cable-force component remains H = 192 kN. The tension magnitude is therefore sqrt(192² + 48²), which is approximately 197.9 kN. Hence Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
The span and UDL of the 24 m cable remain unchanged, but its midspan sag is increased from 3 m to 6 m. What are the new horizontal component H and support tension approximately?
Explanation:
For fixed span and UDL, H is inversely proportional to the sag f. Doubling the sag from 3 m to 6 m therefore halves H from 192 kN to 96 kN. The support vertical component remains 96 kN because the total vertical load has not changed. The new support tension is sqrt(96² + 96²) = approximately 135.8 kN. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
For geometrically similar loading conditions, a parabolic cable carries a UDL w per horizontal metre and has fixed sag f. If its span is doubled while w and f remain unchanged, by what factor does the horizontal component of tension change?
Explanation:
The horizontal component for this loading is H = wL²/(8f). With w and f fixed, H is proportional to the square of span L. Doubling L therefore multiplies L² by four. The required horizontal tension component consequently becomes four times its previous value. Therefore Option D is correct.
వివరణ:
Question 12
ప్రశ్న 12
A perfectly flexible cable is supported at the same level over a span of 20 m. A single 80 kN vertical load acts at midspan, where the cable sag is 4 m. Neglect cable self-weight. What are the horizontal component H and the tension in either cable segment?
Explanation:
The corresponding simply supported beam has a midspan moment PL/4 = 80 × 20/4 = 400 kN·m. The cable theorem gives Hf = 400, so H = 100 kN. By symmetry, each vertical support reaction is 40 kN. The cable segment tension is sqrt(100² + 40²) = approximately 107.7 kN. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A flexible cable with supports at the same level spans 20 m. A 60 kN point load acts 8 m from the left support, and the cable dip at the load point is 3 m. Neglect self-weight. What are the vertical reactions and horizontal component H?
Explanation:
The vertical reactions are the same as those of a simply supported beam under the same point load. Thus VB = 60(8/20) = 24 kN and VA = 36 kN. The beam moment directly under the load is VA × 8 = 288 kN·m. The cable theorem gives H(3) = 288, so H = 96 kN. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
For the cable in the previous question, what are the tension magnitudes in the left and right cable segments?
Explanation:
The horizontal component of tension is H = 96 kN in both segments. The left segment must also provide a vertical component equal to VA = 36 kN. Its tension is therefore sqrt(96² + 36²) = approximately 102.5 kN. The right segment has vertical component 24 kN and tension sqrt(96² + 24²) = approximately 99.0 kN. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
A perfectly flexible cable carries only a number of isolated vertical point loads between its supports, and cable self-weight is neglected. What is the cable geometry between adjacent load points?
Explanation:
No distributed transverse load acts between two adjacent concentrated-load points. The cable force within such an unloaded interval therefore has a constant direction as well as a constant magnitude. Consequently the cable centre-line is straight over that interval. A concentrated load causes equilibrium of the adjoining segment tensions and creates a change or kink in slope. Therefore Option D is correct.
వివరణ:
Question 16
ప్రశ్న 16
A flexible cable spans 20 m between supports at the same level. It carries 30 kN at x = 5 m and 50 kN at x = 12 m from the left support. The dip at x = 12 m is 4 m. Neglect self-weight. What are VA, VB and H?
Explanation:
Taking moments about the left support gives VB(20) = 30(5) + 50(12). Hence VB = 37.5 kN and VA = 80 - 37.5 = 42.5 kN. The corresponding beam moment at x = 12 m is 42.5(12) - 30(7) = 300 kN·m. Since the cable dip there is 4 m, H = 300/4 = 75 kN. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the two-point-load cable in the previous question, what is the cable dip at x = 5 m?
Explanation:
The beam moment at x = 5 m is VA × 5 because the first point load acts at that location. Thus M0 = 42.5 × 5 = 212.5 kN·m. The horizontal component of cable tension has already been found as 75 kN. The general cable theorem gives y = M0/H = 212.5/75 = approximately 2.833 m. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
For the same 20 m cable carrying 30 kN at x = 5 m and 50 kN at x = 12 m, with H = 75 kN, which cable segment has the greatest tension and what is its approximate magnitude?
Explanation:
The left segment has horizontal component 75 kN and vertical component equal to VA = 42.5 kN. Its tension is therefore sqrt(75² + 42.5²) = approximately 86.2 kN. The middle-segment vertical component has magnitude 12.5 kN, giving approximately 76.0 kN tension. The right segment has a vertical component of 37.5 kN and tension approximately 83.9 kN, so the left segment governs. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
For a symmetric parabolic cable with supports at the same level carrying a UDL per unit horizontal span, where do the minimum and maximum cable tensions occur?
Explanation:
The horizontal component H remains constant throughout the cable. At midspan the slope is zero, so the vertical component of cable tension is zero. The tension there is therefore simply H and is the minimum value. The absolute cable slope is largest at the supports, where the vertical component is also largest. Consequently the maximum tension occurs at the supports and Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
A symmetric parabolic cable has span L, midspan sag f and carries a UDL w per unit horizontal span. Which pair correctly gives the tangent slope magnitude at a support and the corresponding support tension?
Explanation:
Differentiating y = 4fx(L-x)/L² gives a support slope magnitude of 4f/L. The support vertical reaction under the symmetric UDL is wL/2. The horizontal component of cable tension remains H. The support tension is therefore the vector magnitude sqrt[H² + (wL/2)²]. Hence Option D is correct.
వివరణ:
Question 21
ప్రశ్న 21
A parabolic cable spans 40 m between equal-level supports and carries 5 kN/m uniformly over the horizontal span. What midspan sag is required if the horizontal component of cable tension is to be 250 kN?
Explanation:
For this loading, H = wL²/(8f). Rearranging gives f = wL²/(8H). Substitution gives f = 5 × 40²/(8 × 250). The numerator is 8000 and the denominator is 2000, producing f = 4 m. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
A parabolic cable spans 40 m, has a midspan sag of 4 m and carries 5 kN/m uniformly over the horizontal span. What is the maximum cable tension approximately?
Explanation:
The horizontal component is H = 5 × 40²/(8 × 4) = 250 kN. Each vertical support reaction is wL/2 = 100 kN. The maximum tension occurs at either support. Its magnitude is sqrt(250² + 100²) = approximately 269.3 kN. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
What is the principal introductory structural purpose of a stiffening girder or stiffened deck in a conventional suspension bridge?
Explanation:
A highly flexible main cable tends to change shape when concentrated live loads move across the span. A stiffening girder or sufficiently stiff deck distributes those loads over a larger region of the suspension system. This reduces undesirable local distortion and improves the roadway profile. The main cable nevertheless remains fundamentally a tension member rather than being converted into a bending beam. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which statement correctly relates ideal funicular cable and arch behaviour under the same system of vertical loads?
Explanation:
A funicular form is a geometry for which the applied loading can be carried without bending in the idealized system. A flexible cable naturally develops such a form while carrying axial tension. If the corresponding geometry is inverted and adequately restrained, an arch can carry the same type of loading primarily through axial compression. This tension-compression duality is an important conceptual relationship between cables and arches. Therefore Option D is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about suspension cables:
1. An ideal flexible cable cannot resist bending moment or axial compression.
2. Under a UDL uniform per unit horizontal span, an equal-level cable assumes a parabolic profile.
3. For such a parabolic cable, the maximum tension occurs at midspan because the sag is maximum there.
4. In a suspension bridge, hangers transfer deck loading to the main cable and a stiffening girder can help distribute moving loads.
Which statements are correct?
1. An ideal flexible cable cannot resist bending moment or axial compression.
2. Under a UDL uniform per unit horizontal span, an equal-level cable assumes a parabolic profile.
3. For such a parabolic cable, the maximum tension occurs at midspan because the sag is maximum there.
4. In a suspension bridge, hangers transfer deck loading to the main cable and a stiffening girder can help distribute moving loads.
Which statements are correct?
Explanation:
Statement 1 is correct because the ideal flexible cable is a tension-only structural element with negligible flexural resistance. Statement 2 is correct for a UDL measured per unit horizontal projection. Statement 3 is incorrect because the tangent is horizontal at midspan, making the tension there equal to the minimum value H, while the support tensions are larger. Statement 4 correctly describes the basic load-transfer and stiffening functions in a suspension bridge. Therefore Statements 1, 2 and 4 are correct and Option C is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. It resists axial tension only, and the tension at every point acts tangentially to the cable. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. A cable carries primarily axial tension, a funicular arch carries primarily axial compression, and a beam commonly carries bending moment and shear. సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: D. Hangers transfer deck loads to the main cables, towers support the cables, and anchorages resist the cable end pull. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. A cable under uniform load per unit horizontal span forms a parabola, while a cable hanging under its own uniformly distributed weight per unit cable length forms a catenary. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. Hy = M0 సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. y = 4fx(L-x)/L² and H = wL²/(8f) సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. Vertical reaction = 96 kN each and H = 192 kN సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. Minimum = 192 kN at midspan; maximum = 214.7 kN at the supports సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: C. y = 2.25 m and T ≈ 197.9 kN సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. H = 96 kN and support tension ≈ 135.8 kN సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. It becomes four times. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. H = 100 kN and T ≈ 107.7 kN సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. VA = 36 kN, VB = 24 kN and H = 96 kN సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: A. Left segment ≈ 102.5 kN; right segment ≈ 99.0 kN సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. A straight segment between adjacent concentrated-load points, with changes in slope occurring at the loads సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: B. VA = 42.5 kN, VB = 37.5 kN and H = 75 kN సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 2.833 m సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. Left support to x = 5 m; approximately 86.2 kN సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. Minimum at midspan and maximum at the supports సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: D. tanθ = 4f/L and T = sqrt[H² + (wL/2)²] సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: C. 4 m సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: A. 269.3 kN సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: B. To distribute concentrated or moving live loads over a greater length and reduce excessive local deformation of the flexible cable-deck system సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. Their funicular forms are geometrically corresponding inverted shapes, with the cable carrying tension and the arch carrying compression. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: C. 1, 2 and 4 only సమాధానం: C.
