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Question 1
ప్రశ్న 1
A triangular distributed load varies linearly from zero at end A to 12 kN/m at end B over a length of 6 m. What are the magnitude and location of its equivalent concentrated load measured from A?
Explanation:
The equivalent load equals the area of the triangular load diagram. Its magnitude is 0.5 × 6 × 12 = 36 kN. The centroid of a triangular load lies at two-thirds of the loaded length measured from the zero-intensity end. Hence its location is 2 × 6/3 = 4 m from A. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
A trapezoidal distributed load varies linearly from 4 kN/m at A to 10 kN/m at B over a 6 m length. What are the magnitude and approximate location of its resultant measured from A?
Explanation:
The resultant magnitude equals the area of the trapezoid, which is (4 + 10) × 6/2 = 42 kN. The load can be decomposed into a 4 kN/m rectangle and a triangular addition rising from zero to 6 kN/m. The rectangle contributes 24 kN at 3 m, while the triangle contributes 18 kN at 4 m from A. The resultant location is therefore (24 × 3 + 18 × 4)/42 = approximately 3.43 m from A. Hence Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
An isolated concentrated couple is applied at an interior point of a beam, and there is no concentrated transverse force at that same point. Which local SFD/BMD behaviour is correct?
Explanation:
A concentrated couple has moment but no resultant transverse force. It therefore does not cause the shear-force jump associated with a concentrated vertical load. It does, however, introduce a finite discontinuity in the bending-moment diagram. The magnitude of the BMD jump equals the magnitude of the applied couple, while its direction depends on the sign convention and couple sense. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A beam is supported at A and B, 6 m apart, and overhangs 2 m beyond B to C. A downward point load of 20 kN acts at C. Taking upward reactions as positive and sagging moment as positive, what are RA, RB and the bending moment at B?
Explanation:
Taking moments about A gives RB × 6 = 20 × 8, so RB = 26.67 kN upward. Vertical equilibrium then gives RA = 20 - 26.67 = -6.67 kN. The overhang load causes a hogging moment at B of magnitude 20 × 2 = 40 kN·m. With sagging taken as positive, this moment is -40 kN·m. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
A beam is simply supported at A and B, 6 m apart, and has a 2 m overhang BC. A UDL of 10 kN/m acts only over the overhang BC. What are the reactions at A and B and the bending moment at B, taking sagging as positive?
Explanation:
The overhang UDL has a resultant of 10 × 2 = 20 kN acting 1 m beyond B, or 7 m from A. Taking moments about A gives RB × 6 = 20 × 7, so RB = 23.33 kN. Vertical equilibrium gives RA = 20 - 23.33 = -3.33 kN. The moment at B due to the overhang load is -wL²/2 = -10 × 2²/2 = -20 kN·m. Hence Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
A beam has 2 m overhangs at both ends, with supports A and B located 6 m apart. A 10 kN downward load acts at the left free end and a 20 kN downward load acts at the right free end. What are the upward reactions RA and RB?
Explanation:
The total downward load is 30 kN, so RA + RB = 30 kN. Taking moments about support A, the left overhang load produces a moment opposite to that of the right overhang load. Equilibrium gives RB × 6 + 10 × 2 - 20 × 8 = 0. Solving gives RB = 23.33 kN and hence RA = 6.67 kN. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
A simply supported beam of span 8 m carries a triangular UVL varying from zero at the left support to 12 kN/m at the right support. At what distance from the left support does the maximum bending moment occur, and what is its approximate value?
Explanation:
The triangular load resultant is 0.5 × 8 × 12 = 48 kN acting 5.333 m from the left support. The reactions are therefore RA = 16 kN and RB = 32 kN. Since w(x) = 1.5x, the shear is V(x) = 16 - 0.75x², which becomes zero at x = 4.619 m. The bending moment there is M = 16x - 0.25x³ = approximately 49.27 kN·m. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
A simply supported beam of span 6 m carries a triangular UVL varying from 15 kN/m at the left support to zero at the right support. What is the approximate location and magnitude of the maximum bending moment?
Explanation:
The triangular load has a resultant of 45 kN acting 2 m from the left support. The support reactions are RA = 30 kN and RB = 15 kN. With w(x) = 15 - 2.5x, the shear equation is V = 30 - 15x + 1.25x². The relevant zero-shear root within the span is x = approximately 2.536 m, and the corresponding bending moment is approximately 34.64 kN·m. Hence Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
A simply supported beam 6 m long carries a trapezoidal load varying linearly from 4 kN/m at the left support to 10 kN/m at the right support. What is the approximate location and value of the maximum bending moment?
Explanation:
The total load is 42 kN and its resultant acts approximately 3.43 m from the left support. Equilibrium gives RA = 18 kN and RB = 24 kN. The load intensity is w(x) = 4 + x, so the shear is V = 18 - 4x - x²/2. Setting the shear equal to zero gives x = approximately 3.211 m, and the bending moment there is approximately 31.66 kN·m. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
A cantilever of length 6 m carries a triangular load varying from 12 kN/m at the fixed end to zero at the free end. What are the magnitudes of the fixed-end shear force and bending moment?
Explanation:
The triangular load resultant is 0.5 × 6 × 12 = 36 kN. For a triangle whose maximum intensity occurs at the fixed end, the resultant acts one-third of the length from that end. Its lever arm about the fixed support is therefore 2 m. The fixed-end shear magnitude is 36 kN and the moment magnitude is 36 × 2 = 72 kN·m. Hence Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
A cantilever of length 6 m carries a triangular load varying from zero at the fixed end to 12 kN/m at the free end. What are the magnitudes of the fixed-end shear force and bending moment?
Explanation:
The triangular load resultant is again 0.5 × 6 × 12 = 36 kN. Since the intensity is zero at the fixed end and maximum at the free end, the resultant acts two-thirds of the span from the fixed end. The lever arm is therefore 4 m. The fixed-end bending-moment magnitude is 36 × 4 = 144 kN·m. Thus Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
A beam is supported at A and B, 6 m apart, and overhangs 2 m to C. A UDL of 6 kN/m acts over AB and an 18 kN downward point load acts at C. Taking sagging moment as positive, where is the interior point of contraflexure in AB and what is the maximum positive bending moment?
Explanation:
The UDL over AB contributes 36 kN at 3 m from A, and the overhang point load contributes 18 kN at 8 m from A. Equilibrium gives RB = 42 kN and RA = 12 kN. Within AB, M(x) = 12x - 3x², so its non-end zero occurs at x = 4 m. Shear is V = 12 - 6x, so maximum positive moment occurs at x = 2 m and equals 12 kN·m. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A simply supported beam 6 m long is subjected only to a clockwise concentrated couple of 30 kN·m at an interior point. Taking upward reactions as positive, what are RA and RB?
Explanation:
A pure applied couple contributes no net vertical force. Therefore vertical equilibrium requires RA + RB = 0. Taking moments about the left support, the right reaction must provide a 30 kN·m counteracting moment. Hence RB × 6 = 30, giving RB = +5 kN and RA = -5 kN. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A simply supported beam 8 m long carries a UDL of 4 kN/m over the full span and a clockwise applied couple of 16 kN·m at midspan. Taking sagging bending moment as positive, what is the maximum positive bending moment?
Explanation:
Whole-beam equilibrium gives RA = 14 kN and RB = 18 kN. Immediately to the left of midspan, the bending moment is 14 × 4 - 4 × 4²/2 = 24 kN·m. Under the stated sagging-positive convention, the clockwise couple produces a positive BMD jump of 16 kN·m at its location. The bending moment immediately to the right is therefore 24 + 16 = 40 kN·m, after which it decreases toward zero at the right support. Hence Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A distributed load varies according to w(x) = kx over a beam interval, where k is constant and no concentrated loads or couples occur inside that interval. What are the general polynomial forms of the shear-force and bending-moment variations?
Explanation:
A load intensity proportional to x is a linearly varying distributed load. Since the derivative of shear is proportional to the negative of load intensity, integration of a linear load gives a quadratic shear variation. Bending moment is obtained by integrating the shear function. Integrating a quadratic shear function produces a cubic bending-moment function. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Over a 4 m beam interval, a triangular downward load varies from zero to 8 kN/m. If the shear-force ordinate at the beginning of the interval is +10 kN, what is the shear-force ordinate at the end?
Explanation:
The decrease in shear across a loaded interval equals the resultant downward load. The triangular load resultant is 0.5 × 4 × 8 = 16 kN. Starting with a shear ordinate of +10 kN, the final shear becomes 10 - 16 = -6 kN. This result is equivalent to integrating the load intensity over the interval. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
Over a 5 m beam interval, the shear-force diagram varies linearly from +20 kN at the left end to -10 kN at the right end. Using dM/dx = V, what is the net change in bending moment across the interval?
Explanation:
The change in bending moment equals the algebraic area under the shear-force diagram. Because the shear varies linearly, its average value is [20 + (-10)]/2 = 5 kN. Multiplying by the interval length gives ΔM = 5 × 5 = 25 kN·m. The positive algebraic area means the bending moment at the right end is 25 kN·m greater than at the left end. Hence Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Why can the bending moment at a simple support located inside a continuous overhanging beam be non-zero even though a pin or roller support itself does not resist an external moment reaction?
Explanation:
A pin or roller support does not supply an external reaction couple. However, the beam is physically continuous through an intermediate support unless an internal hinge is specifically provided there. Loads on the overhang can therefore create an internal bending moment in the beam section at that support. The support provides the required force reaction while the beam itself transmits the internal moment across the section. Therefore Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A trapezoidal downward load varies linearly from 6 kN/m at the left end to 18 kN/m at the right end over 8 m. What are the resultant magnitude and its location measured from the left end?
Explanation:
The trapezoidal load magnitude is (6 + 18) × 8/2 = 96 kN. It can be decomposed into a 6 kN/m rectangle of 48 kN and an additional triangle of 48 kN. The rectangle acts 4 m from the left end, while the triangle acts at 2 × 8/3 = 5.333 m from the zero end of its triangular component. The combined centroid is (48 × 4 + 48 × 5.333)/96 = approximately 4.67 m. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
A cantilever 4 m long carries a trapezoidal downward load varying from 4 kN/m at the fixed end to 10 kN/m at the free end. What are the magnitudes of the fixed-end shear force and bending moment?
Explanation:
The total trapezoidal load is (4 + 10) × 4/2 = 28 kN. Decomposing it gives a 16 kN rectangular load acting 2 m from the fixed end and a 12 kN triangular load acting 2.667 m from the fixed end. The total fixed-end moment magnitude is therefore 16 × 2 + 12 × 2.667 = 64 kN·m. The fixed-end shear-force magnitude equals the total load of 28 kN. Hence Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
A beam is supported at A and B, 5 m apart, and overhangs 2 m beyond B to C. A 20 kN downward point load acts 3 m from A and a 10 kN downward point load acts at C. Taking sagging moment as positive, which pair correctly gives the bending moment at B and the interior point of contraflexure between the 20 kN load and B?
Explanation:
Moment equilibrium gives RB × 5 = 20 × 3 + 10 × 7, so RB = 26 kN and RA = 4 kN. The overhang point load produces a bending moment at B of -10 × 2 = -20 kN·m. Between x = 3 m and x = 5 m, the bending moment is M = 4x - 20(x - 3) = 60 - 16x. Setting this expression equal to zero gives x = 3.75 m from A. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
The bending-moment ordinates immediately on the two sides of an interior point differ in magnitude by 25 kN·m, while the shear-force diagram has no jump at that point. Which localized external action is consistent with this observation?
Explanation:
A finite jump in bending moment indicates the presence of a concentrated applied couple. The magnitude of that jump equals the magnitude of the couple, subject to sign convention for its direction. Because the shear-force diagram has no jump, there is no concentrated transverse force at that location. A distributed load changes shear continuously rather than creating a finite jump at a single point. Therefore Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
Consider the following statements about advanced SFD and BMD construction:
1. The resultant of a triangular distributed load acts at one-third of its base measured from the higher-intensity end.
2. An applied concentrated couple produces a jump in the bending-moment diagram but no corresponding concentrated-force jump in the shear-force diagram.
3. A point of contraflexure requires the bending moment to pass through zero and change sign.
4. For a linearly varying distributed load, shear varies linearly and bending moment varies parabolically.
Which statements are correct?
1. The resultant of a triangular distributed load acts at one-third of its base measured from the higher-intensity end.
2. An applied concentrated couple produces a jump in the bending-moment diagram but no corresponding concentrated-force jump in the shear-force diagram.
3. A point of contraflexure requires the bending moment to pass through zero and change sign.
4. For a linearly varying distributed load, shear varies linearly and bending moment varies parabolically.
Which statements are correct?
Explanation:
Statement 1 is correct because the centroid of a triangular load lies one-third of the base from the larger-intensity end. Statement 2 is correct because a pure couple changes moment directly without introducing a transverse force resultant. Statement 3 is correct because contraflexure requires a reversal in bending-moment sign and therefore curvature sense. Statement 4 is incorrect because a linearly varying load produces a quadratic shear-force variation and a cubic bending-moment variation. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 24
ప్రశ్న 24
Match each loading feature with the corresponding SFD/BMD implication:
1. Triangular UVL
2. Trapezoidal UVL
3. Concentrated couple
4. Loaded overhang beyond a simple support
P. May create a non-zero hogging internal moment at the adjacent support
Q. Can be decomposed conveniently into rectangular and triangular load components
R. Produces a quadratic shear variation and cubic bending-moment variation over the loaded region
S. Produces a finite bending-moment jump without a transverse-force jump
Which matching is correct?
1. Triangular UVL
2. Trapezoidal UVL
3. Concentrated couple
4. Loaded overhang beyond a simple support
P. May create a non-zero hogging internal moment at the adjacent support
Q. Can be decomposed conveniently into rectangular and triangular load components
R. Produces a quadratic shear variation and cubic bending-moment variation over the loaded region
S. Produces a finite bending-moment jump without a transverse-force jump
Which matching is correct?
Explanation:
A triangular UVL has load intensity varying linearly with position, so its shear is quadratic and its moment is cubic, making item 1 match R. A trapezoidal UVL can be decomposed into a rectangle plus a triangle, so item 2 matches Q. A concentrated couple creates a finite BMD jump without a corresponding shear-force jump, so item 3 matches S. A loaded overhang can generate a hogging bending moment at the adjacent intermediate simple support, so item 4 matches P. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
A beam is supported at A and B, 6 m apart, and overhangs 2 m to C. A triangular distributed load acts over AB, varying from zero at A to 9 kN/m at B, and a 12 kN downward point load acts at C. Taking sagging moment as positive, which pair correctly gives the bending moment at B and the interior point of contraflexure in AB?
Explanation:
The triangular load over AB has a resultant of 27 kN acting 4 m from A. Whole-beam equilibrium gives RB = 34 kN and RA = 5 kN. The overhang point load gives the moment at B as -12 × 2 = -24 kN·m, which is consistent with the span-side calculation. Within AB, the bending moment is M(x) = 5x - 0.25x³, so the non-end zero occurs at x = √20 = approximately 4.47 m. Therefore Option B is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. 36 kN acting 4 m from A సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. 42 kN at 3.43 m సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. The bending-moment diagram has a finite jump equal in magnitude to the applied couple, while the shear-force diagram has no corresponding force jump. సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: B. RA = -6.67 kN, RB = +26.67 kN and MB = -40 kN·m సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. RA = -3.33 kN, RB = +23.33 kN and MB = -20 kN·m సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: B. RA = 6.67 kN and RB = 23.33 kN సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. 4.62 m and 49.27 kN·m సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: A. 2.54 m from the left support and 34.64 kN·m సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. 3.21 m and 31.66 kN·m సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. 36 kN and 72 kN·m సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. 36 kN and 144 kN·m సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: B. Contraflexure at 4 m from A; maximum positive moment 12 kN·m సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: B. RA = -5 kN and RB = +5 kN సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 40 kN·m immediately to the right of the applied couple సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. Shear is quadratic and bending moment is cubic. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: C. -6 kN సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. +25 kN·m సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: B. Because the beam section over the support can transmit internal bending moment generated by loads on the overhang, even though the support reaction itself has no moment component. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. 96 kN at approximately 4.67 m సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. 28 kN and 64 kN·m సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. MB = -20 kN·m and contraflexure at x = 3.75 m from A సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. A 25 kN·m concentrated couple సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: B. 1, 2 and 3 only సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. 1-R, 2-Q, 3-S, 4-P సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: B. MB = -24 kN·m and contraflexure at approximately 4.47 m from A సమాధానం: B.
