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Question 1
ప్రశ్న 1
While constructing shear-force and bending-moment diagrams for a statically determinate beam, which procedure is fundamentally correct?
Explanation:
Support reactions should first be obtained from the static equilibrium equations for the entire beam. The beam can then be cut at appropriate sections to determine internal shear force and bending moment. The variation between key sections follows from the nature of the distributed loading and the load-shear-moment relationships. This procedure avoids relying only on memorized diagram shapes. Therefore Option B gives the correct analytical sequence.
వివరణ:
Question 2
ప్రశ్న 2
A simply supported beam of span 8 m carries a 40 kN point load at midspan. What are the support reactions and the maximum bending moment?
Explanation:
The load and geometry are symmetric about the beam centre. Therefore the two support reactions are equal and each is 40/2 = 20 kN. The shear changes from +20 kN to -20 kN at the central point load, so the maximum bending moment occurs at midspan. The maximum moment is 20 × 4 = 80 kN·m. Hence Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
A simply supported beam AB of span 6 m carries a 30 kN point load 2 m from support A. What are the vertical reactions and the maximum bending moment?
Explanation:
Taking moments about A gives RB × 6 = 30 × 2. Therefore RB = 10 kN and vertical equilibrium gives RA = 20 kN. The shear is positive between A and the load and negative immediately after the load. The maximum bending moment consequently occurs beneath the point load and equals 20 × 2 = 40 kN·m. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A simply supported beam 5 m long carries a UDL of 12 kN/m over its entire span. What are the support reactions and maximum bending moment?
Explanation:
The total uniformly distributed load is 12 × 5 = 60 kN. Because the loading is symmetric, each support reaction is 30 kN. For a simply supported beam carrying a full-span UDL, zero shear occurs at midspan. The maximum bending moment is wL²/8 = 12 × 5²/8 = 37.5 kN·m. Hence Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A simply supported beam of span 8 m carries a UDL of 6 kN/m over the entire span and a 20 kN downward point load 2 m from the left support. What are the support reactions?
Explanation:
The full-span UDL has a resultant of 6 × 8 = 48 kN acting 4 m from the left support. Taking moments about the left support gives RB × 8 = 48 × 4 + 20 × 2. This gives RB = 29 kN. Vertical equilibrium then gives RA = 48 + 20 - 29 = 39 kN. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
A simply supported beam of span 10 m carries a UDL of 4 kN/m over its full span and a 30 kN point load at 6 m from the left support. Where does the maximum bending moment occur and what is its value?
Explanation:
The total UDL is 40 kN and equilibrium gives RA = 32 kN and RB = 38 kN. Immediately to the left of the point load, the shear is 32 - 4 × 6 = +8 kN. Immediately to the right, the 30 kN load changes the shear to -22 kN, so the bending moment has a local maximum at that load point. The bending moment there is 32 × 6 - 4 × 6²/2 = 120 kN·m. Hence Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
A simply supported beam of span 8 m carries a UDL of 10 kN/m over the left-hand 4 m only. At what distance from the left support does the maximum bending moment occur, and what is its value?
Explanation:
The UDL resultant is 40 kN acting 2 m from the left support. Moment equilibrium gives RB = 10 kN and hence RA = 30 kN. Within the loaded region, shear is V = 30 - 10x, so it becomes zero at x = 3 m. The bending moment there is 30 × 3 - 10 × 3²/2 = 45 kN·m. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
A simply supported beam of span 8 m carries a UDL of 10 kN/m over the right-hand 4 m only. At what distance from the left support does the maximum bending moment occur, and what is its value?
Explanation:
The UDL resultant is 40 kN and acts 6 m from the left support. Equilibrium gives RA = 10 kN and RB = 30 kN. Between x = 4 m and x = 8 m, the shear is V = 10 - 10(x - 4), which becomes zero at x = 5 m. The bending moment there is 10 × 5 - 10 × 1²/2 = 45 kN·m. Hence Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
A cantilever beam 3 m long carries a downward point load of 18 kN at its free end. Taking sagging bending moment as positive, which pair correctly gives the internal shear-force magnitude and the fixed-end bending moment?
Explanation:
Every section between the free-end load and the fixed support must transmit the same 18 kN transverse force. The shear-force magnitude is therefore constant at 18 kN over the cantilever. The bending-moment magnitude increases linearly toward the fixed end. At the support its magnitude is PL = 18 × 3 = 54 kN·m. The bending is hogging and therefore negative under the stated convention, so Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
A cantilever beam 4 m long carries a UDL of 7 kN/m over the entire span. Taking sagging bending moment as positive, what are the fixed-end shear-force magnitude and bending moment?
Explanation:
The total UDL is wL = 7 × 4 = 28 kN. Therefore the fixed support carries a vertical reaction of 28 kN and the fixed-end internal shear has this magnitude. The UDL resultant acts at the midpoint, 2 m from the support. The fixed-end bending-moment magnitude is 28 × 2 = 56 kN·m. It is a hogging moment and is therefore -56 kN·m under the stated convention, making Option B correct.
వివరణ:
Question 11
ప్రశ్న 11
A cantilever 5 m long carries a UDL of 4 kN/m over its entire length and a 20 kN downward point load located 3 m from the fixed end. What are the vertical fixed-support reaction magnitude and fixed-end bending-moment magnitude?
Explanation:
The UDL produces a total downward load of 4 × 5 = 20 kN. Including the concentrated load, the total downward force is 20 + 20 = 40 kN, so the vertical reaction magnitude is 40 kN. The UDL contributes a fixed-end moment of 20 × 2.5 = 50 kN·m. The point load contributes 20 × 3 = 60 kN·m, giving a total moment magnitude of 110 kN·m. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
A cantilever beam is 6 m long and carries a UDL of 5 kN/m only over the outermost 2 m adjacent to the free end. What are the fixed-end shear-force magnitude and bending-moment magnitude?
Explanation:
The loaded length is 2 m, so the resultant UDL is 5 × 2 = 10 kN. The loaded region extends from 4 m to 6 m from the fixed support, so its resultant acts at x = 5 m. The fixed-end shear magnitude equals the total transverse load and is therefore 10 kN. The fixed-end bending-moment magnitude is 10 × 5 = 50 kN·m. Hence Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A cantilever beam 6 m long carries a UDL only over the first 2 m measured from the fixed support, while the remaining 4 m up to the free end is unloaded. What are the shear force and bending moment at a section 3 m from the fixed support?
Explanation:
The section at 3 m lies beyond the entire loaded portion when viewed toward the free end. The free-body segment from that section to the free end therefore contains no transverse load or applied couple. Equilibrium of that free segment requires the internal shear force at the cut to be zero. It also requires the internal bending moment at the cut to be zero. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
A simply supported beam of span 9 m carries a 20 kN point load at x = 3 m and a 30 kN point load at x = 6 m from the left support. What is the maximum bending moment?
Explanation:
Moment equilibrium about the left support gives RB × 9 = 20 × 3 + 30 × 6. Thus RB = 26.67 kN and RA = 23.33 kN. The shear is +23.33 kN before the first load, +3.33 kN between the two loads and -26.67 kN after the second load. The bending moment is therefore maximum at x = 6 m and equals 23.33 × 6 - 20 × 3 = 80 kN·m. Hence Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A simply supported beam 8 m long carries two equal point loads of 25 kN located 2 m and 6 m from the left support. What is the structural response between the two loads?
Explanation:
Symmetry gives support reactions of 25 kN each. Immediately after the first 25 kN point load, the shear reduces from +25 kN to zero. It remains zero until the second point load is reached because there is no load between the two point loads. Since the moment slope equals shear under a consistent sign convention, zero shear means the bending moment is constant in this region. Its value is 25 × 2 = 50 kN·m, so Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
A simply supported beam 6 m long carries point loads of 15 kN at 1.5 m and 25 kN at 4.5 m from the left support. What is the maximum bending moment?
Explanation:
Taking moments about the left support gives RB × 6 = 15 × 1.5 + 25 × 4.5. This gives RB = 22.5 kN and RA = 17.5 kN. The shear remains positive at +2.5 kN between the loads and becomes negative after the 25 kN load. The maximum bending moment therefore occurs at x = 4.5 m and equals 17.5 × 4.5 - 15 × 3 = 33.75 kN·m. Hence Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A simply supported beam 10 m long carries a UDL of 8 kN/m over its entire span. Define the shear-force ordinate V at a section as the algebraic sum of vertical forces to the left of that section, taking upward forces as positive, and take sagging bending moment as positive. What are V and M at a section 3 m from the left support?
Explanation:
The total UDL is 8 × 10 = 80 kN, so symmetry gives each support reaction as 40 kN upward. At x = 3 m, the downward load to the left of the section is 8 × 3 = 24 kN. Under the explicitly stated convention, the shear-force ordinate is V = 40 - 24 = +16 kN. The sagging bending moment is M = 40 × 3 - 8 × 3²/2 = 84 kN·m. Therefore Option A is uniquely correct.
వివరణ:
Question 18
ప్రశ్న 18
A simply supported beam of span 12 m carries a UDL of 6 kN/m over the full span. At what section is the shear force zero, and what is the bending moment there?
Explanation:
The total load is 6 × 12 = 72 kN, so each reaction is 36 kN. At a distance x from the left support, the shear force is V = 36 - 6x. Setting V equal to zero gives x = 6 m, which is the midspan. The bending moment there is 36 × 6 - 6 × 6²/2 = 108 kN·m. Hence Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
Over a 6 m beam interval, the shear force decreases linearly from +24 kN to -12 kN. Adopt dV/dx = -w and dM/dx = V, with downward load intensity w taken as positive. What uniform load intensity acts over the interval and what is the net change in bending moment?
Explanation:
The shear-force slope is (-12 - 24)/6 = -6 kN/m. Since dV/dx = -w, the downward UDL intensity is therefore 6 kN/m. The change in bending moment equals the algebraic area under the shear-force diagram. That area is the average shear, [24 + (-12)]/2 = 6 kN, multiplied by 6 m, giving +36 kN·m. Thus Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
At a concentrated downward point load, the shear-force ordinate changes from +14 kN immediately to the left of the load to -26 kN immediately to its right. What is the magnitude of the point load, assuming there is no support reaction at that same point?
Explanation:
A concentrated transverse point load causes a finite jump in the shear-force diagram. The jump here is -26 - (+14) = -40 kN. The negative jump corresponds to a downward point load under the stated interpretation. Its magnitude is therefore 40 kN. The bending-moment diagram remains continuous at that point if no concentrated couple is also applied, so Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
Over a 2 m beam segment, the shear force is constant at +15 kN and the bending moment at the left end of the segment is +20 kN·m. Using dM/dx = V, what is the bending moment at the right end?
Explanation:
The change in bending moment equals the algebraic area under the shear-force diagram. A constant shear of +15 kN over 2 m gives a moment increase of 15 × 2 = 30 kN·m. The initial bending moment is +20 kN·m. Therefore the final bending moment is 20 + 30 = 50 kN·m. Hence Option D is correct.
వివరణ:
Question 22
ప్రశ్న 22
The same downward point load P is applied first at the free end of a cantilever of length L and then at a point L/2 from the fixed support. How do the fixed-end shear-force magnitude and fixed-end bending-moment magnitude compare between the two cases?
Explanation:
The fixed support must balance the same transverse load P in both loading positions. Therefore the fixed-end shear-force magnitude remains P. The fixed-end bending-moment magnitude equals the load multiplied by its distance from the fixed support. Moving the load from L to L/2 changes that moment from PL to PL/2. Therefore the shear is unchanged while the moment is halved, making Option B correct.
వివరణ:
Question 23
ప్రశ్న 23
A simply supported beam 6 m long carries a UDL of 4 kN/m over its full span together with an 18 kN downward point load at midspan. What is the maximum bending moment?
Explanation:
The full-span UDL has a resultant of 24 kN and the total applied load is 24 + 18 = 42 kN. Symmetry gives each support reaction as 21 kN. Immediately to the left of midspan the shear is 21 - 4 × 3 = +9 kN, while immediately to the right the point load changes it to -9 kN. The maximum moment therefore occurs at midspan and equals 21 × 3 - 4 × 3²/2 = 45 kN·m. Hence Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements for beams carrying only point loads and uniformly distributed loads:
1. Between concentrated loads, a region having no distributed load has constant shear force.
2. Under a constant UDL, the shear-force diagram is linear and the bending-moment diagram is parabolic.
3. A point load causes a finite jump in bending moment even when no concentrated couple is applied there.
4. A zero-shear point at which shear changes sign corresponds to a local extremum of bending moment.
Which statements are correct?
1. Between concentrated loads, a region having no distributed load has constant shear force.
2. Under a constant UDL, the shear-force diagram is linear and the bending-moment diagram is parabolic.
3. A point load causes a finite jump in bending moment even when no concentrated couple is applied there.
4. A zero-shear point at which shear changes sign corresponds to a local extremum of bending moment.
Which statements are correct?
Explanation:
Statement 1 is correct because in the absence of distributed loading the shear-force slope is zero. Statement 2 is correct because a constant load intensity gives linear shear, whose integration gives a quadratic bending-moment variation. Statement 3 is incorrect because an ordinary concentrated transverse force causes a shear jump while bending moment remains continuous when no couple is present. Statement 4 is correct because zero shear gives zero BMD slope, and a shear sign change causes that slope to reverse. Therefore Statements 1, 2 and 4 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each beam-loading case with its characteristic shear-force and bending-moment behaviour:
1. Simply supported beam with a single central point load
2. Simply supported beam with a full-span UDL
3. Cantilever with a free-end point load
4. Cantilever with a full-span UDL
P. Constant shear and linearly varying bending moment over the span
Q. Linear shear and parabolic bending moment, both reaching their largest magnitudes at the fixed end
R. Shear is piecewise constant with a jump at midspan and bending moment is piecewise linear with its maximum at midspan
S. Shear varies linearly across the span and bending moment is parabolic with its maximum at midspan
Which matching is correct?
1. Simply supported beam with a single central point load
2. Simply supported beam with a full-span UDL
3. Cantilever with a free-end point load
4. Cantilever with a full-span UDL
P. Constant shear and linearly varying bending moment over the span
Q. Linear shear and parabolic bending moment, both reaching their largest magnitudes at the fixed end
R. Shear is piecewise constant with a jump at midspan and bending moment is piecewise linear with its maximum at midspan
S. Shear varies linearly across the span and bending moment is parabolic with its maximum at midspan
Which matching is correct?
Explanation:
A central point load on a simply supported beam produces piecewise constant shear and a triangular, piecewise linear bending-moment diagram, so item 1 matches R. A full-span UDL on a simply supported beam produces linear shear and a parabolic moment diagram with the maximum moment at midspan, so item 2 matches S. A cantilever with a free-end point load has constant shear and a linearly varying bending moment, so item 3 matches P. A full-span UDL on a cantilever produces linear shear and a parabolic bending-moment diagram whose largest magnitudes occur at the fixed end, so item 4 matches Q. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Determine support reactions from whole-beam equilibrium, evaluate internal shear and moment from section equilibrium, and then connect the ordinates consistently with the loading. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. 20 kN at each support and 80 kN·m సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: B. RA = 20 kN, RB = 10 kN and Mmax = 40 kN·m సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: A. 30 kN at each support and 37.5 kN·m సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: C. RA = 39 kN and RB = 29 kN సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. At the 30 kN load, 6 m from the left support; 120 kN·m సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. 3 m; 45 kN·m సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: C. 5 m; 45 kN·m సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. 18 kN and -54 kN·m సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: B. 28 kN and -56 kN·m సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: A. 40 kN and 110 kN·m సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: B. 10 kN and 50 kN·m సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. Shear = 0 and moment = 0 సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 80 kN·m సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. Shear force is zero and bending moment is constant at 50 kN·m. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 33.75 kN·m సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. V = +16 kN and M = +84 kN·m సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. x = 6 m; M = 108 kN·m సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. w = 6 kN/m and ΔM = 36 kN·m సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 40 kN సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: D. +50 kN·m సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: B. Fixed-end shear is unchanged, while fixed-end moment is halved. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: B. 45 kN·m సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: B. 1, 2 and 4 only సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 1-R, 2-S, 3-P, 4-Q సమాధానం: A.
