Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
At a section of a transversely loaded beam, which statement correctly distinguishes shear force from bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Internal shear force at a beam section represents the transverse force resultant required for equilibrium of either cut portion of the beam. Internal bending moment represents the corresponding moment resultant required for rotational equilibrium. Shear force has dimensions of force and is commonly expressed in kN. Bending moment has dimensions of force multiplied by length and is commonly expressed in kN·m. Therefore Option A correctly distinguishes the two beam resultants.
వివరణ:
Question 2
ప్రశ్న 2
Using the common beam convention in which sagging bending moment is positive, which bending-moment condition is negative?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under the commonly used beam sign convention, sagging bending moment is taken as positive. Sagging normally places the upper fibres in compression and the lower fibres in tension. Hogging reverses this curvature and places the upper fibres in tension and lower fibres in compression. Hogging bending moment is therefore assigned a negative sign under this convention. Hence Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which statement most accurately describes the differential relationships among distributed load intensity, shear force and bending moment in a beam?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Differential equilibrium of a beam element establishes direct relationships among load intensity, shear force and bending moment. Distributed loading governs how shear varies along the beam. Shear force in turn governs how bending moment varies along the beam. The algebraic signs in the differential equations depend on the sign convention adopted for load, shear and moment. Therefore Option A states the relationship in a convention-independent and technically correct form.
వివరణ:
Question 4
ప్రశ్న 4
At an isolated downward point load on a beam, with no applied concentrated couple at that same point, what happens to the shear-force and bending-moment diagrams?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A concentrated transverse force produces a sudden change in the internal shear force when the load point is crossed. The magnitude of the shear jump equals the magnitude of the concentrated transverse load, with its sign determined by convention. A finite concentrated force does not by itself produce a finite jump in bending moment. The bending-moment diagram therefore remains continuous at that point when no concentrated couple is applied there. Hence Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
An isolated concentrated couple is applied at an interior point of a beam, with no concentrated transverse force at that point. Which diagram behaviour is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A concentrated couple introduces a finite moment directly at its point of application. The bending-moment diagram consequently exhibits a finite jump whose magnitude corresponds to the applied couple, subject to the adopted sign convention. A pure couple has no resultant transverse force. It therefore does not cause the shear-force jump associated with a concentrated vertical load. Hence Option C correctly describes the diagram behaviour.
వివరణ:
Question 6
ప్రశ్న 6
Over a beam segment containing no distributed load and no concentrated load within the segment, what form do the shear-force and bending-moment diagrams take?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With no distributed load over a segment, the derivative of shear with respect to position is zero under the corresponding differential equilibrium relation. Shear force therefore remains constant throughout that segment. Since the slope of the bending-moment diagram is proportional to shear, a constant shear produces a linear variation of bending moment. The constant shear need not be zero, so the bending moment need not be constant. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
A beam segment carries a constant uniformly distributed load and contains no concentrated loads within the segment. Which qualitative variation is expected?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A constant distributed load gives a constant rate of change of shear force. The shear-force diagram therefore varies linearly over the loaded region. Because the rate of change of bending moment is related to shear, integrating a linear shear variation produces a quadratic or parabolic bending-moment variation. This is a fundamental load-shear-moment relationship rather than merely a memorized diagram shape. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
At an interior point where the shear force passes continuously through zero and changes sign, what is normally true of the bending moment at that point?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The slope of the bending-moment diagram is related directly to the shear force. When shear becomes zero, the slope of the moment diagram becomes zero under a consistent sign convention. If the shear also changes sign through that point, the slope of the moment diagram reverses sign. The bending moment therefore has a local extremum, which may be a maximum or minimum depending on the loading and signs. Hence Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
Which statement correctly defines an interior point of contraflexure in a beam?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A point of contraflexure is an interior point where the bending moment passes through zero and changes sign. The change of moment sign corresponds to a reversal between sagging and hogging curvature in the usual elastic interpretation. Merely having zero bending moment is not sufficient unless a sign change occurs across the point. A simple support may have zero bending moment without being called an interior point of contraflexure. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Between two beam sections, what does the algebraic area under the shear-force diagram represent, using a sign convention in which dM/dx = V?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under the stated convention, the differential relationship is dM/dx = V. Integrating this equation between two sections gives the change in bending moment as the integral of shear force over the distance. Graphically, that integral is the algebraic area under the shear-force diagram. Positive and negative shear areas contribute with their respective signs. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
A simply supported beam of span 6 m carries a concentrated load of 30 kN at midspan. What are the support reactions and maximum bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The loading and support geometry are symmetric, so the two vertical reactions are equal. Vertical equilibrium gives RA + RB = 30 kN, hence each reaction is 15 kN. The shear changes from +15 kN to -15 kN at the central load, so the bending moment is maximum there. The maximum moment is 15 × 3 = 45 kN·m. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
A simply supported beam of span 8 m carries a uniformly distributed load of 10 kN/m over the entire span. Where does the shear force become zero, and what is the maximum bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The total uniformly distributed load is 10 × 8 = 80 kN and acts at midspan. Symmetry gives reactions of 40 kN at each support. At a distance x from the left support, shear is V = 40 - 10x, so zero shear occurs at x = 4 m. The bending moment there is 40 × 4 - 10 × 4²/2 = 80 kN·m. Hence Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A simply supported beam AB has a span of 6 m and carries a 20 kN point load 2 m from support A. What are the support reactions and the maximum bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Taking moments about A gives RB × 6 = 20 × 2, so RB = 6.67 kN. Vertical equilibrium then gives RA = 20 - 6.67 = 13.33 kN. The shear is positive to the left of the point load and negative immediately to its right, so the bending moment is maximum under the load. The maximum moment is RA × 2 = 13.33 × 2 = approximately 26.67 kN·m. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A simply supported beam of span 10 m carries a 20 kN point load at 3 m from the left support and a 30 kN point load at 7 m from the left support. What is the maximum bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Taking moments about the left support gives RB × 10 = 20 × 3 + 30 × 7 = 270 kN·m, so RB = 27 kN. The left reaction is therefore 50 - 27 = 23 kN. Shear is +23 kN before the first load, +3 kN between the loads and -27 kN after the second load, so the maximum moment occurs at the 30 kN load at x = 7 m. The bending moment there is 23 × 7 - 20 × 4 = 81 kN·m. Hence Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A simply supported beam of span 6 m carries a UDL of 10 kN/m over the left 3 m only. What is the location and value of the maximum bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The UDL resultant is 30 kN acting 1.5 m from the left support. Taking moments about the left support gives RB = 30 × 1.5/6 = 7.5 kN and RA = 22.5 kN. Within the loaded region the shear is V = 22.5 - 10x, which becomes zero at x = 2.25 m. The moment there is 22.5 × 2.25 - 10 × 2.25²/2 = approximately 25.31 kN·m. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A simply supported beam of span 6 m carries a triangular distributed load that varies linearly from zero at support A to 12 kN/m at support B. What is the location of maximum bending moment measured from A, and its approximate magnitude?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The triangular load resultant is 0.5 × 6 × 12 = 36 kN and acts 4 m from A. Moment equilibrium gives RB = 24 kN and RA = 12 kN. The load intensity is w(x) = 2x kN/m, so the shear is V(x) = 12 - x² and becomes zero at x = √12 ≈ 3.46 m. The bending moment is M(x) = 12x - x³/3, which gives approximately 27.71 kN·m at that location. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A cantilever beam of length 3 m carries a downward point load of 12 kN at its free end. Taking sagging bending moment as positive, what are the internal shear-force magnitude and fixed-end bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The cantilever must transmit the 12 kN transverse load through every section between the load and the fixed support. The internal shear-force magnitude is therefore 12 kN throughout the span. The maximum bending-moment magnitude occurs at the fixed end and equals PL = 12 × 3 = 36 kN·m. A downward end load causes hogging at the fixed support, which is negative when sagging is taken as positive. Hence Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
A cantilever beam 4 m long carries a uniformly distributed load of 5 kN/m over its entire length. Taking sagging moment as positive, what are the fixed-end shear-force magnitude and bending moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The total uniformly distributed load is wL = 5 × 4 = 20 kN. The fixed support must provide an equal vertical reaction, so the internal shear magnitude at the fixed end is 20 kN. The UDL resultant acts at L/2 = 2 m from the support. The fixed-end moment magnitude is therefore 20 × 2 = 40 kN·m, equivalent to wL²/2. The moment is hogging and hence negative under the stated convention, making Option A correct.
వివరణ:
Question 19
ప్రశ్న 19
A beam is supported at A and B, 6 m apart, and has a 2 m overhang beyond B to point C. A 20 kN downward point load acts at C. Taking upward reactions as positive, what are the vertical reactions at A and B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Taking moments about A gives RB × 6 = 20 × 8. Therefore RB = 26.67 kN upward. Vertical equilibrium requires RA + RB = 20 kN, giving RA = -6.67 kN. The negative value means that the reaction at A acts downward in the idealized equilibrium solution and a real support would need the ability to resist uplift. Hence Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
A beam is simply supported at A and B, 6 m apart, with a 2 m overhang BC. A UDL of 10 kN/m acts over AB and a 20 kN downward point load acts at free end C. Taking sagging moment as positive, where is the interior point of contraflexure in span AB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The UDL over AB has a resultant of 60 kN at 3 m from A, and the overhang load is 20 kN at 8 m from A. Moment equilibrium gives RB = (60 × 3 + 20 × 8)/6 = 56.67 kN, so RA = 23.33 kN. Within AB, the bending moment is M(x) = 23.33x - 10x²/2. Setting M(x) = 0 gives the non-end solution x = 23.33/5 ≈ 4.67 m, across which the moment changes from sagging to hogging. Therefore Option D is correct.
వివరణ:
Question 21
ప్రశ్న 21
A simply supported beam of span 6 m is subjected only to a clockwise concentrated couple of 24 kN·m at an interior point. Taking upward support reactions as positive, what are the reactions at the left and right supports?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because the applied load is a pure couple, there is no net vertical external force. Thus vertical equilibrium requires RA + RB = 0. Taking moments about the left support, the right reaction must provide an anticlockwise moment of 24 kN·m to balance the clockwise couple. Therefore RB × 6 = 24, giving RB = +4 kN and RA = -4 kN. Hence Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
Between two beam sections 3 m apart, the shear force decreases linearly from +18 kN to 0. Adopt the convention dV/dx = -w and dM/dx = V. If no concentrated load occurs between the sections, what are the uniform downward load intensity and increase in bending moment over this interval?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The shear changes from 18 kN to zero over 3 m, so its slope is (0 - 18)/3 = -6 kN/m. Since dV/dx = -w, the uniform downward load intensity is therefore 6 kN/m. The change in bending moment equals the algebraic area under the shear diagram because dM/dx = V. The triangular area is 0.5 × 3 × 18 = 27 kN·m and is positive. Therefore Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
A beam region has zero shear force throughout a finite length and contains no concentrated couple within that region. What can be concluded about the bending moment there?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The derivative of bending moment with respect to position is proportional to shear force under a consistent beam sign convention. If shear is zero throughout a finite region, the bending-moment slope is zero there. The bending moment is therefore constant over that region. It need not be zero; a non-zero constant moment corresponds to a pure-bending region. Hence Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about shear-force and bending-moment diagrams:
1. A concentrated transverse point load causes a finite jump in the shear-force diagram.
2. A concentrated applied couple causes a finite jump in the bending-moment diagram.
3. If shear passes continuously through zero and changes sign, the bending moment has a local extremum.
4. Every point at which bending moment is zero is necessarily a point of contraflexure.
Which statements are correct?
1. A concentrated transverse point load causes a finite jump in the shear-force diagram.
2. A concentrated applied couple causes a finite jump in the bending-moment diagram.
3. If shear passes continuously through zero and changes sign, the bending moment has a local extremum.
4. Every point at which bending moment is zero is necessarily a point of contraflexure.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because a concentrated transverse load produces a jump in shear equal to its magnitude with sign determined by convention. Statement 2 is correct because an applied concentrated couple produces a finite discontinuity in bending moment. Statement 3 is correct because zero shear corresponds to zero moment slope, and a shear sign change produces a local moment maximum or minimum. Statement 4 is incorrect because a support or beam end may have zero moment without the bending moment changing sign through that point. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match the loading condition with its characteristic local effect on shear-force and bending-moment diagrams:
1. Region with no distributed load
2. Constant UDL
3. Concentrated transverse load
4. Concentrated applied couple
P. Linear shear variation and parabolic moment variation
Q. Finite shear jump with continuous bending moment when no couple acts there
R. Constant shear and linear bending moment
S. Bending-moment jump without a corresponding transverse-force jump
Which matching is correct?
1. Region with no distributed load
2. Constant UDL
3. Concentrated transverse load
4. Concentrated applied couple
P. Linear shear variation and parabolic moment variation
Q. Finite shear jump with continuous bending moment when no couple acts there
R. Constant shear and linear bending moment
S. Bending-moment jump without a corresponding transverse-force jump
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With no distributed load, shear remains constant and bending moment varies linearly, so item 1 matches R. A constant UDL produces a linear shear variation and a parabolic bending-moment variation, so item 2 matches P. A concentrated transverse force produces a finite jump in shear while the moment remains continuous if no couple acts there, so item 3 matches Q. A concentrated couple changes bending moment discontinuously without introducing a transverse-force jump, so item 4 matches S. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Shear force is the algebraic resultant of transverse forces on one side of the section, while bending moment is the algebraic resultant moment of those loads about the section. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. Hogging curvature with tension at the top fibres సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. The rate of change of shear is related to distributed load intensity, and the rate of change of bending moment is related to shear force, with algebraic signs depending on the adopted convention. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. Shear force has a finite jump equal to the point-load magnitude, while bending moment remains continuous. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. The bending-moment diagram has a finite jump corresponding to the applied couple, while shear does not have a force jump. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. Shear remains constant and bending moment varies linearly. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: A. Shear varies linearly and bending moment varies parabolically. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: B. The bending moment has a local maximum or minimum. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: B. A point where bending moment changes sign and the beam curvature reverses sense సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. The change in bending moment between the sections సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. 15 kN at each support and 45 kN·m సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: B. At 4 m from the left support; 80 kN·m సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: B. RA = 13.33 kN, RB = 6.67 kN, Mmax = 26.67 kN·m సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 81 kN·m సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. At 2.25 m from the left support; approximately 25.31 kN·m సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. x ≈ 3.46 m; Mmax ≈ 27.71 kN·m సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. Shear 12 kN and fixed-end moment -36 kN·m సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. 20 kN and -40 kN·m సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. RA = -6.67 kN and RB = +26.67 kN సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: D. Approximately 4.67 m from A సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. RA = -4 kN and RB = +4 kN సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: B. w = 6 kN/m and ΔM = 27 kN·m సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: B. The bending moment is constant throughout the region and may be non-zero. సమాధానం: B.
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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-R, 2-P, 3-Q, 4-S సమాధానం: A.